JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(f : [0,1]\,\to R\) be such that \(f\,(xy) = f\,(x)\,f\,(y)\) for all \(x,y\,\in [0,1],\) and \(f \,(0)\,\ne 0.\) If \(y=y\,(x)\) satisfies the differential equation,\(\frac{{dy}}{{dx}} = f(x)\) with \(y(0) = 1,\) then \(y\left( {\frac{1}{4}} \right) + y\left( {\frac{3}{4}} \right)\) is equal to
- A \(4\)
- B \(3\)
- C \(5\)
- D \(2\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\(f\left( {xy} \right) = f\left( x \right).f\left( y \right)\) \(f\left( 0 \right) = 1\) as \(f\left( 0 \right) \ne 0\) \( \Rightarrow f\left( x \right) = 1\) \(\frac{{dy}}{{dx}} = f\left( x \right) = 1\) \( \Rightarrow y = x + c\) At, \(x = 0,y = 1 \Rightarrow c = 1\)…
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